MCQ elimination and the maths of negative marking
When guessing pays off, and how to remove options without solving fully.
Expected value of a guess
With +4 for a correct answer and −1 for a wrong one across 4 options, a blind guess is worth 0.25×4 − 0.75×1 = +0.25. That is barely positive, and the variance is high.
Remove one option and it becomes (1/3)×4 − (2/3)×1 = +0.67. Remove two and it is +1.5.
| Marking | Options left | EV of guess |
|---|---|---|
| +4 / −1 | 4 | +0.25 |
| +4 / −1 | 3 | +0.67 |
| +4 / −1 | 2 | +1.50 |
| +3 / −1 (CAT) | 4 | 0.00 |
| +3 / −1 (CAT) | 2 | +1.00 |
So in CAT a blind guess is worth nothing. Guess only after you've eliminated at least one option.
Elimination tools
- Units and dimensions. Throw out options with the wrong dimensions.
- Limiting cases. Set a variable to 0 or ∞. Which options still behave sensibly?
- Sign and direction. Does the quantity increase or decrease? (For example, convex lens power is positive.)
- Trap patterns. Setters include the answer you get from the most common slip. In guided mode the trap steps show these explicitly.
- Order of magnitude. Estimate roughly before calculating precisely.